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Find the average value of the function f(x)=(6)/(9-4x) from x=3 to x=9. Express your answer as a constant times ln 3. Answer: ◻ln 3

Find the average value of the function f(x)=69−4x f(x)=\frac{6}{9-4 x} from x=3 x=3 to x=9 x=9 . Express your answer as a constant times ln⁡3 \ln 3 .\newlineAnswer: □ln⁡3\square \ln 3

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Q. Find the average value of the function f(x)=69−4x f(x)=\frac{6}{9-4 x} from x=3 x=3 to x=9 x=9 . Express your answer as a constant times ln⁡3 \ln 3 .\newlineAnswer: □ln⁡3\square \ln 3
  1. Calculate Interval Width: To find the average value of a continuous function f(x)f(x) on the interval [a,b][a, b], we use the formula for the average value of a function on an interval, which is given by:\newlineAverage value = 1(b−a)∫abf(x) dx\frac{1}{(b-a)} \int_{a}^{b} f(x) \, dx\newlineHere, a=3a = 3 and b=9b = 9, and f(x)=6(9−4x)f(x) = \frac{6}{(9-4x)}.
  2. Set Up Integral: First, we calculate b−ab - a, which is the width of the interval.\newlineb−a=9−3=6b - a = 9 - 3 = 6
  3. Perform u-Substitution: Next, we set up the integral to find the average value.\newlineAverage value = (1/6)×∫3969−4x dx(1/6) \times \int_{3}^{9} \frac{6}{9-4x} \, dx
  4. Rewrite Integral in u: To integrate f(x)=69−4xf(x) = \frac{6}{9-4x}, we perform a u-substitution.\newlineLet u=9−4xu = 9 - 4x, then du=−4dxdu = -4 dx, or dx=−du4dx = -\frac{du}{4}.\newlineWhen x=3x = 3, u=9−4(3)=9−12=−3u = 9 - 4(3) = 9 - 12 = -3.\newlineWhen x=9x = 9, u=9−4(9)=9−36=−27u = 9 - 4(9) = 9 - 36 = -27.
  5. Simplify Integral: Now we rewrite the integral in terms of uu.\newlineAverage value = 16\frac{1}{6} * ∫−3−276u∗(−14)du\int_{-3}^{-27} \frac{6}{u} * \left(-\frac{1}{4}\right) du
  6. Evaluate Integral: We can simplify the integral by taking constants out.\newlineAverage value = (16)×(−64)×∫−3−27(1u)du(\frac{1}{6}) \times (\frac{-6}{4}) \times \int_{-3}^{-27} (\frac{1}{u}) du\newlineAverage value = (−14)×∫−3−27(1u)du(-\frac{1}{4}) \times \int_{-3}^{-27} (\frac{1}{u}) du
  7. Remove Absolute Value: The integral of 1u\frac{1}{u} du is ln⁡∣u∣\ln|u|.\newlineAverage value = −14\frac{-1}{4} * (ln⁡∣−27∣\ln|-27| - ln⁡∣−3∣\ln|-3|)
  8. Apply Logarithm Property: Since the natural logarithm function ln⁡∣x∣\ln|x| is the same for positive and negative values of xx, we can remove the absolute value signs.\newlineAverage value = (−1/4)×(ln⁡(27)−ln⁡(3))(-1/4) \times (\ln(27) - \ln(3))
  9. Simplify Natural Log: We use the property of logarithms that ln⁡(a)−ln⁡(b)=ln⁡(ab)\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right).\newlineAverage value = (−14)⋅ln⁡(273)\left(-\frac{1}{4}\right) \cdot \ln\left(\frac{27}{3}\right)\newlineAverage value = (−14)⋅ln⁡(9)\left(-\frac{1}{4}\right) \cdot \ln(9)
  10. Correct Limits of Integration: We know that ln⁡(9)\ln(9) is the same as 2×ln⁡(3)2 \times \ln(3).\newlineAverage value = (−14)×2×ln⁡(3)(-\frac{1}{4}) \times 2 \times \ln(3)\newlineAverage value = (−12)×ln⁡(3)(-\frac{1}{2}) \times \ln(3)
  11. Correct Limits of Integration: We know that ln⁡(9)\ln(9) is the same as 2×ln⁡(3)2 \times \ln(3).
    Average value = (−1/4)×2×ln⁡(3)(-1/4) \times 2 \times \ln(3)
    Average value = (−1/2)×ln⁡(3)(-1/2) \times \ln(3)Correcting the limits of integration after the u-substitution, we should have:
    Average value = (1/6)×∫−27−3(6)/u×(−1/4) du(1/6) \times \int_{-27}^{-3} (6)/u \times (-1/4) \, du
    This changes the sign of the integral because the limits are now in the correct order.
    Average value = (1/6)×(6/4)×∫−27−3(1/u) du(1/6) \times (6/4) \times \int_{-27}^{-3} (1/u) \, du
    Average value = (1/4)×(ln⁡∣−3∣−ln⁡∣−27∣)(1/4) \times (\ln|-3| - \ln|-27|)
  12. Correct Limits of Integration: We know that ln⁡(9)\ln(9) is the same as 2×ln⁡(3)2 \times \ln(3).
    Average value = (−1/4)×2×ln⁡(3)(-1/4) \times 2 \times \ln(3)
    Average value = (−1/2)×ln⁡(3)(-1/2) \times \ln(3)Correcting the limits of integration after the u-substitution, we should have:
    Average value = (1/6)×∫−27−3(6)/u×(−1/4) du(1/6) \times \int_{-27}^{-3} (6)/u \times (-1/4) \, du
    This changes the sign of the integral because the limits are now in the correct order.
    Average value = (1/6)×(6/4)×∫−27−3(1/u) du(1/6) \times (6/4) \times \int_{-27}^{-3} (1/u) \, du
    Average value = (1/4)×(ln⁡∣−3∣−ln⁡∣−27∣)(1/4) \times (\ln|-3| - \ln|-27|)Removing the absolute value signs and using the property of logarithms:
    Average value = (1/4)×(ln⁡(3)−ln⁡(27))(1/4) \times (\ln(3) - \ln(27))
    Average value = (1/4)×ln⁡(3/27)(1/4) \times \ln(3/27)
    Average value = (1/4)×ln⁡(1/9)(1/4) \times \ln(1/9)
  13. Correct Limits of Integration: We know that ln⁡(9)\ln(9) is the same as 2×ln⁡(3)2 \times \ln(3).
    Average value = (−1/4)×2×ln⁡(3)(-1/4) \times 2 \times \ln(3)
    Average value = (−1/2)×ln⁡(3)(-1/2) \times \ln(3)Correcting the limits of integration after the u-substitution, we should have:
    Average value = (1/6)×∫−27−3(6)/u×(−1/4) du(1/6) \times \int_{-27}^{-3} (6)/u \times (-1/4) \, du
    This changes the sign of the integral because the limits are now in the correct order.
    Average value = (1/6)×(6/4)×∫−27−3(1/u) du(1/6) \times (6/4) \times \int_{-27}^{-3} (1/u) \, du
    Average value = (1/4)×(ln⁡∣−3∣−ln⁡∣−27∣)(1/4) \times (\ln|-3| - \ln|-27|)Removing the absolute value signs and using the property of logarithms:
    Average value = (1/4)×(ln⁡(3)−ln⁡(27))(1/4) \times (\ln(3) - \ln(27))
    Average value = (1/4)×ln⁡(3/27)(1/4) \times \ln(3/27)
    Average value = (1/4)×ln⁡(1/9)(1/4) \times \ln(1/9)Since 2×ln⁡(3)2 \times \ln(3)00 is the same as 2×ln⁡(3)2 \times \ln(3)11, we have:
    Average value = 2×ln⁡(3)2 \times \ln(3)22
    Average value = (−1/2)×ln⁡(3)(-1/2) \times \ln(3)
    However, we need to correct the sign due to the earlier mistake. The average value should be positive, so we have:
    Average value = 2×ln⁡(3)2 \times \ln(3)44

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